klastiesym

klastiesym

Answered

2022-10-24

Find the Laplace transform of the function H ( t a ) t n . That's what I have done so far:
L { H ( t a ) t n } = 0 e s t H ( t a ) t n d t = a e s t t n d t = 0 e s ( t + a ) ( t + a ) n d t
Could you tell me how I can continue?

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t5an1izqp

t5an1izqp

Expert

2022-10-25Added 13 answers

You can obtain directly from the integral
L { H ( t a ) t n } = a e s t t n d t = 1 s n + 1 a s e z z n d z = 1 s n + 1 Γ ( n + 1 , a s )
recalling the integral expression of incomplete Gamma function for integer n
Γ ( n , x ) = x e z z n 1 d z .
NOTE Another useful methodology may be the following.
Observing that
H ( t a ) t n = H ( t a ) ( t a + a ) n = H ( t a ) k = 0 n ( n k ) ( t a ) k a n k
the Laplace transform becomes
L { H ( t a ) t n } = k = 0 n ( n k ) a n k L { H ( t a ) ( t a ) k }
recalling that L { H ( t a ) f ( t a ) } = e a s F ( s ). Using L { t n } = n ! s n + 1 , we obtain
L { H ( t a ) t n } = e a s k = 0 n ( n k ) a n k k ! s k + 1 = n ! e a s s n + 1 k = 0 n ( a s ) n k ( n k ) ! = 1 s n + 1 n ! e a s ν = 0 n ( a s ) ν ν !
and recalling that
Γ ( n , z ) = ( n 1 ) ! e z ν = 0 n 1 z ν ν !
is the incomplete Gamma function for integer n, we obtain finally
L { H ( t a ) t n } = 1 s n + 1 Γ ( n + 1 , a s ) .

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Pellagra3d

Pellagra3d

Expert

2022-10-26Added 1 answers

You know that L { U ( t a ) } = e a s s , a > 0. And
L { t n f ( t ) } = ( 1 ) n ( L ( f ( t ) ) ) ( n )
Now let think about ( e a s s ) ( n ) . Using Maple, I found:
n = 1 ( e a s s ) = exp ( a s ) a s + 1 s 2 n = 2 ( e a s s ) = exp ( a s ) a 2 s 2 + 2 a s + 2 s 3 n = 3 ( e a s s ) = exp ( a s ) a 3 s 3 + 3 a 2 s 2 + 6 a s + 6 s 4
And so I think the possible pattern could be
( 1 ) n k = 0 n n ! ( n k ) ! s n k a n k s n + 1

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